At what depth below the surface of the earth acceleration due to gravity $'g'$ will be half of its value at $1600\ km$ above the surface of the earth?

Variation of $g$ outside the earth

${ g }^{ 1 }=g\left( 1-\dfrac { 2h }{ R } \right)$

${ g }^{ 1 }=10\left( 1-\dfrac { 2\times 1600 }{ 6400 } \right)$

${ g }^{ 1 }=5m/{ s }^{ 2 }$

Variation of $g$ inside the earth

${ g }^{ 11 }=g\left( 1-\dfrac { h }{ R } \right)$

Now, it is given that,

${ g }^{ 11 }=\dfrac { { g }^{ 1 } }{ 2 }$

$2.5m/{ s }^{ 2 }=10\left( 1-\dfrac { h }{ 6400 } \right)$

$\dfrac { 1 }{ 4 } =1-\dfrac { h }{ 6400 }$

$h=\dfrac { 3 }{ 4 } \times 6400$ Km

$h=4.8\times { 10 }^{ 6 }$ m